4/26/2004, STOKES THEOREM II Math21a, O. KnillHOMEWORK. Section: 13.3: 34,38, p.1138: 30,32,34curl: curl(M, N, P ) = (Py− Nz, Mz− Px, Nx− My).gradient: grad(f) = (fx, fy, fz).Fundamental theorem of line integralscurve r(t) : [a, b] → C with boundary {r(a), r(b)}.Zbagrad(f)(r(t)) · r0(t) dt = f(r(b)) − f(r(a))Stokes theorem:surface r(u, v) : R → S with boundary r(t) : [a, b] → CZ ZRcurl(F )(r(u, v)) · (ru× rv)dudv =ZbaF (r(t)) · r0(t) dtREMINDERS:CURL:curl(F ) = ∇ × FGRAD: grad(f) = ∇fFTL:RCgrad(f) · dr = f(B) − f(A)STOKES:R RScurl(F ) · dS =RCF · drCURL(GRAD)=0: ∇ × ∇f =~0CONSERVATIVE FIELDS. F is conservative if is a gradient field F = ∇f. The fundamental theorem of lineintegral implies that the line integral along closed curves is zero and that line integrals are path independent.We have also seen that F = grad(f) implies that F has zero curl. If we know that F is conservative, how do wecompute f? If F = (M, N, P ) = (fx, fy, fz), we got f by integration. There is an other method which you doin the homework in two dimensions: to get the potential value f, find a path CPfrom the origin to the pointP = (x, y, z) and computeRCF · dr. Because line integrals are path independent, the fundamental theorem ofline integrals givesRCF · dr = f (P ).CONNECTED. A region is called connected if one can connect any two points in the region with a path.SIMPLY CONNECTED. A region is called simply connected if it is connected and every path in the regioncan be deformed to a point within the region.EXAMPLES 2D:Simply connected. Not simply connected. Simply connected. Not simply connected.EXAMPLES 3D:A solid ball is simply con-nected.A solid torus is not simplyconnected.The complement of a solidball is simply connected.The complement of a solidtorus is not simply con-nected.THEOREM.In a simply connected region D, a vector field F is conservative ifand only if curl(F ) =~0 everywhere inside D.Proof. We already know that F = ∇f implies curl(F ) =~0. Toshow the converse, we verify that the line integral along any closedcurve C in D is zero. This is equivalent to the path independenceand allows the construction of the potential f with F = ∇f.By assumption, we can deform the curve to a point: if r0(t) isthe original curve and r1(t) is the curve r1(t) = P which stays atone point, define a parametrized surface S by r(t, s) = rs(t). Byassumption, curl(F ) =~0 and therefore the flux of curl(F ) throughS is zero. By Stokes theorem, the line integral along the boundaryC of the surface S is zero too.THE NASH PROBLEM. Nash challenged his multivariable class in the movie ”A beautiful mind” with aproblem, where the region is not simply connected.Find a region X of R3with the property that if V is the set ofvector fields F on R3\ X which satisfy curl(F ) = 0 and W is theset of vector fields F which are conservative: F = ∇f. Then, thespace V /W should be 8 dimensional.You solve this problem as an inclass exercise (ICE). The problemis to find a region D in space, in which one can find 8 differentclosed paths Ciso that for every choice of constants (c1, ..., c8),one can find a vector field F which has zero curl in D and forwhich one hasRC1F dr = c1, ...,RC8F dr = c8.One of the many solutions is cut out 8 tori from space. For eachtorus, there is a vector field Fi(a vortex ring), which has itsvorticity located inside the ring and such that the line integral ofa path which winds once around the ring is 1. The vector fieldF = c1F1+ ... + c8F8has the required properties.CLOSED SURFACES. Surfaces with no boundaries are called closed surfaces. For example, the surface of adonought, or the surface of a sphere are closed surfaces. A half sphere is not closed, its boundary is a circle.Half a doughnut is not closed. Its boundary consists of two circles.THE ONE MILLION DOLLAR QUESTION. One of the Millenium problems is to determine whether any threedimensional space which is simply connected is deformable to a sphere. This is called the Poincare conjecture.LINEINTEGRAL IN HIGHER DIMENSIONS. Line integrals are defined in the same way in higher dimensions.RCF · dr, where · is the dot product in d dimensions and dr = r0(t)dt.CURL IN HIGHER DIMENSIONS. In d-dimensions, the curl is the field curl(F )ij= ∂xjFi− ∂xiFjwithd2components. In 4 dimensions, it has 6 components. In 2 dimensions it has 1 component, in 3 dimensions, ithas 3 components.SURFACE INTEGRAL IN HIGHER DIMENSIONS. In d dimensions, a surface element in the ij-plane iswritten as dSij. The flux integral of the curl of F through S is defined asR Rcurl(F ) · dS, where the dotproduct isPi<jcurl(F )ijdSij. If S is given by a map X from a domain R in the 2-plane to Rd, U = ∂uX andV = ∂vX are tangent vectors to that plane and dSij(u, v) = (UiVj− UjVi) dudv.STOKES THEOREM IN HIGHER DIMENSIONS. If S is a two dimensional surface in d-dimensional spaceand C is its boundary, thenR RScurl(F ) · dS =RCF ·
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